Sequence And Series Question 243
Question: The interior angles of a polygon are in A.P. If the smallest angle be $ 120^{o} $ and the common difference be 5o, then the number of sides is
[IIT 1980]
Options:
A) 8
B) 10
C) 9
D) 6
Show Answer
Answer:
Correct Answer: C
Solution:
Let the number of sides of the polygon be $ n $ . Then the sum of interior angles of the polygon $ =(2n-4)\frac{\pi }{2}=(n-2)\pi $ Since the angles are in A.P. and $ a=120^{o},\ d=5 $ , therefore $ \frac{n}{2}[2\times 120+(n-1)5]=(n-2)180 $
Þ $ n^{2}-25n+144=0 $
Þ $ (n-9)(n-16)=0 $
Þ $ n=9,\ 16 $ But $ n=16 $ gives $ T_{16}=a+15d=120^{o}+{{15.5}^{o}}=195^{o} $ , which is impossible as interior angle cannot be greater than $ 180^{o} $ . Hence $ n=9 $ .
BETA
